2 Answers
2

Take $$a_n=(-1)^n \frac{1}{n}$$ It is known that $$\sum_{n=1}^\infty (-1)^n \frac{1}{n}=-\log(2),$$ but $$\sum_{n=1}^\infty (-1)^n (-1)^n \frac{1}{n}=\sum_{n=1}^\infty \frac{1}{n}$$ diverges. If you have $A_n\geq 0$ for nearly all $n\in \mathbb{N}$ you have the absolute konvergence and then the statement is true